Find the `pH` of solution prepared by mixing `25 ml` of a `0.5 M` solution of `HCl, 10 ml` of a `0.5 M` solution of `NaOH` and `15 ml` of water

Explanation

We know that for HCl and NaOH, m.e.=m.m.
`{:(,"m.e. of HCl",=0.5 xx 25=12.5),(,"m.e. of NaOH",=0.5 xx 10=5.0):}}`
m.e. of HCl in the resultant mixture =12.5 -5.0=7.5
Total volume =(25+10+15)mL=50mL.
normality of HCl `=(m.e.)/("vol (mL)")=(7.50)/(50)`
molarity `=(7.5)/(50)`
`[H^+]=[HCl]=(7.5)/(50)`
`pH =-log""(7.5)/(50)=0.8239`